Answer:
y=11/3
Step-by-step explanation:
3y=2+9
Divide both sides by 3.
3y/3 = (2+9)/3
y=2/3+9/3
Combine like terms.
<u>y=11/3</u>
Answer:
1. the range of f^-1(x) is {10, 20, 30}.
2. the graph of f^-1(x) will include the point (0, 3)
3. n = 8
Step-by-step explanation:
1. The domain of a function is the range of its inverse, and vice versa. The range of f^-1(x) is {10, 20, 30}.
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2. See above. The domain and range are swapped between a function and its inverse. That means function point (3, 0) will correspond to inverse function point (0, 3).
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3. The n-th term of an arithmetic sequence is given by ...
an = a1 +d(n -1)
You are given a1 = 2, a12 = 211, so ...
211 = 2 + d(12 -1)
209/11 = d = 19 . . . . . solve the above equation for the common difference
Now, we can use the same equation to find n for an = 135.
135 = 2 + 19(n -1)
133/19 = n -1 . . . . . . . subtract 2, divide by 19
7 +1 = n = 8 . . . . . . . . add 1
135 is the 8th term of the sequence.
Answer:
The commission would be $ 920.
Step-by-step explanation:
Given,
Sales in 1997 = $ 9500,
Since, there is a commision of 8 % on the first $ 7500 of sales, 16% on the next $ 7500 of sales, and 20% on sales over $ 15,000,
If x represents the total sales,
Commission for x ≤ 7500 = 8% of x = 0.08x,
Commission for 7500 < x ≤ 15000 = 8% of 7500 + 16% of (x-7500)
= 600 + 0.16(x-7500),
Commission for 7500 < x ≤ 15000 = 8% of 7500 + 16% of 7500 + 20% of (x-15000)
= 600 + 1200 + 0.20(x-15000)
= 1800 + 0.20(x-15000)
Thus, the function that shows the given situation,

Since, 9500 lies between 7500 and 15000,
Therefore, the commission would be,
f(9500)= 600 + 0.16(9500 - 7500) = 600 + 320 = $ 920
Answer:
A. 0.2222
B. Repeat indefinitely
Step-by-step explanation:
A. We have that Scott has 9 pieces in total and he eats 2 of them. That means he eats 2 <em>out of </em>the 9, so we have 2 ÷ 9. (see attachment)
The decimal form is: 0.2222.
B. Clearly, we can see that throughout the long division, we keep getting 20 - 18 = 2, which makes another 20 and so on. So, this will repeat indefinitely.